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icang [17]
4 years ago
6

Helppppppppppppppppppppppppppppp

Mathematics
1 answer:
sukhopar [10]4 years ago
6 0

Answer:

30 degress

Step-by-step explanation:

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In a random sample of 810 women employees, it is found that 81 would prefer working for a female boss. The width of the 95% conf
Nostrana [21]

Answer:

D) ± .0207

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The width of the interval is given by the following formula:

W = z\sqrt{\frac{\pi(1-\pi)}{n}}

In a random sample of 810 women employees, it is found that 81 would prefer working for a female boss.

This means that n = 810, p = \frac{81}{810} = 0.1

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The width of the 95% confidence interval for the proportion of women who prefer a female boss is

W = z\sqrt{\frac{\pi(1-\pi)}{n}}

W = 1.94\sqrt{\frac{0.1*0.9}{810}} = 0.0207

So the correct answer is:

D) ± .0207

4 0
4 years ago
Can you help me find the test statistic AND Pvalue for a hypothesis test. The question is: In a survey respondents were asked wo
vitfil [10]

Answer:

Test statistic z=0.65

P-value=0.52

Fail to reject the null hypothesis.

Step-by-step explanation:

Here we have to perform a hypothesis test on the difference of proportions. We may want to answer if there is significant difference in the proportions of male and females.

The null and alternative hypothesis are:

H_0: \pi_1=\pi_2\\\\H_1: \pi_1\neq\pi_2

The significance level is defined as 0.05.

The YES proportion for females is:

p_1=33/98=0.337

The YES proportion for males is:

p_2=30/102=0.294

The weighted average of p can be calculated as:

p=\frac{n_1*p_1+n_2*p_2}{n_1+n_2}=\frac{33+30}{98+102}=0.315

With this p, we estimate the standard deviation

s=\sqrt{\frac{p(1-p)}{n_1}+\frac{p(1-p)}{n_2} }=\sqrt{\frac{0.315(1-0.315)}{98}+\frac{0.315(1-0.315)}{102} }=0.066

The test statistic z can be calculated as:

z=\frac{p_1-p_2}{s} =\frac{0.337-0.294}{0.066}=\frac{0.043}{0.066}=0.65

We can calculate the p-value for z, taking into account is a two-sided test

P(|z|>0.65)=0.52

The P-value (0.52) is greater than the significance level (0.05), so the effect is not significant. It failed to reject the null hypothesis.

We have enough evidence to conclude that the proportions that vote YES are different between genres.

8 0
4 years ago
Pls help for brainlis
kodGreya [7K]

Answer:

C) 3.96

Step-by-step explanation:

6 0
3 years ago
Mark has a standard deck of 52 cards and a fair two-sided coin. What is the probability that he will pull a jack from the deck o
murzikaleks [220]
0.0025 would be the chance.
7 0
4 years ago
Use substitution to determine which of the following points is a solution to the standard form equation below 2y=5
valkas [14]
Y=2.5 and if u need me to explain i will be glad to 

8 0
4 years ago
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